### All AP Calculus AB Resources

## Example Questions

### Example Question #191 : Ap Calculus Ab

Is the following piecewise function continuous for all ? If not, state where it is discontinuous.

**Possible Answers:**

Yes. The function is continuous at all .

No. The function is not continuous at both and .

No. The function is not continuous at .

No. The function is not continuous at .

No. There are sharp turns at and .

**Correct answer:**

Yes. The function is continuous at all .

To check if the piecewise function is continuous, all we need to do is check that the values at 3 and 5 line up.

At , this means checking that and have the same value. More formally, we are checking to see that , as to be continuous at a point, a function's left and right limits must both equal the function value at that point.

Plugging 3 into both, we see that both of them are 12 at . Thus, they meet up smoothly.

Next, for , we have and . Plugging in 5, we get 22 for both equations.

As all three equations are polynomials, we know they will be continuous everywhere else, and they meet up smoothly at the piecewise bounds, thus ensuring that the function is continuous everywhere.

Note, there *are *sharp turns at and , but this only means the function isn't differentiable at these points -- we're only concerned with continuity, which is if the equations meet up. Thus, the function is continuous.

### Example Question #21 : Functions, Graphs, And Limits

when and

when

At the funciton described above is:

**Possible Answers:**

neither differentiable or continuous

both continuous and diffentiable

undefined

continuous but not differentiable

differentiable but not continuous

**Correct answer:**

both continuous and diffentiable

The answer is both.

If graphed the student will see that the two graphs are continuous at . There is no gap in the graph or no uneven transitions. If the graph is continuous then it is differentiable so it must be both.

### Example Question #1 : An Intuitive Understanding Of Continuity

Which of the following functions contains a removeable discontinuity?

**Possible Answers:**

**Correct answer:**

A removeable discontinuity occurs whenever there is a hole in a graph that could be fixed (or "removed") by filling in a single point. Put another way, if there is a removeable discontinuity at , then the limit as approaches exists, but the value of does not.

For example, the function contains a removeable discontinuity at . Notice that we could simplify as follows:

, where .

Thus, we could say that .

As we can see, the limit of exists at , even though is undefined.

What this means is that will look just like the parabola with the equation EXCEPT when, where there will be a hole in the graph. However, if we were to just define , then we could essentially "remove" this discontinuity. Therefore, we can say that there is a removeable discontinuty at .

The functions

, and

have discontinuities, but these discontinuities occur as vertical asymptotes, not holes, and thus are not considered removeable.

The functions

and are continuous over all the real values of ; they have no discontinuities of any kind.

The answer is

.

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